On the maximum principle for the Riesz transform

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by  Vladimir Eiderman

Dedicated to a memory of remarkable mathematician and man Victor Petrovich Havin
ON THE MAXIMUM PRINCIPLE FOR THE RIESZ TRANSFORM
VLADIMIR EIDERMAN AND FEDOR NAZAROV
Abstract. Let µ be a measure in Rd with compact support and continuous density, and
let
y−x
7 Rsµ(x)= dµ(y), x,y ∈Rd, 0<s<d.
Z |y−x|s+1
1
0 We consider the following conjecture:
2 sup |Rsµ(x)|≤C sup |Rsµ(x)|, C =C(d,s).
n x∈Rd x∈suppµ
a This relationwas knownfor d−1≤s<d, and is still an open problem in the generalcase.
J
We prove the maximum principle for 0<s<1, and also for 0<s<d in the case of radial

7
measure. Moreover,we show that this conjecture is incorrect for non-positive measures.
1

]
A
C 1. Introduction
.
h Let µ be a non-negative finite Borel measure with compact support in Rd, and let 0 < s <
t
a d. The truncated Riesz operator Rs is defined by the equality
m µ,ε
[ y −x
Rs f(x) = f(y)dµ(y), x,y ∈ Rd, f ∈ L2(µ), ε > 0.
1 µ,ε Z |y −x|s+1
|y−x|>ε
v
0 For every ε > 0 the operator Rs is bounded on L2(µ). By Rs we denote a linear operator
µ,ε µ
0 on L2(µ) such that
5
y −x
4 Rsf(x) = f(y)dµ(y),
0 µ Z |y −x|s+1
.
1 whenever the integral exists in the sense of the principal value. We say that Rs is bounded
0 µ
7 on L2(µ) if
1 kRsk := supkRs k < ∞.
: µ µ,ε L2(µ)→L2(µ)
v ε>0
Xi In the case f ≡ 1 the function Rs1(x) is said to be the s-Riesz transform (potential) of µ
µ
r and is denoted by Rsµ(x). If µ has continuous density with respect to the Lebesgue measure
a m in Rd, that is if dµ(x) = ρ(x)dm (x) with ρ(x) ∈ C(Rd), then Rsµ(x) exists for every
d d
x ∈ Rd.
By C,c, possibly with indexes, we denote various constants which may depend only on d
and s.
We consider the following well-known conjecture.
Conjecture 1.1. Let µ be a nonnegative finite Borel measure with compact support and
continuous density with respect to the Lebesgue measure in Rd. There is a constant C such
that
sup |Rsµ(x)| ≤ C sup |Rsµ(x)|. (1.1)
x∈Rd x∈suppµ
For s = d−1 the proof is simple. Obviously,
Rsµ(x) = ∇Us(x), (1.2)
µ
1
2 VLADIMIREIDERMANAND FEDOR NAZAROV
where
1 dµ(y)
Us(x) = , s 6= 1, U1(x) = − log|y −x|dµ(y).
µ s−1 Z |y −x|s−1 µ Z
Thus each component of the vector function Rsµ(x), s = d−1, is harmonic in Rd \suppµ.
Applying the maximum principle for harmonic functions we get (1.1).
For d−1 < s < d, the relation (1.1) was established in [2] under stronger assumption that
ρ ∈ C∞(Rd). In fact it was proved that (1.1) holds for each component of Rsµ with C = 1
as in the case s = d−1. The proof is based on the formula which recovers a density ρ from
Us. But this method does not work for s < d−1.
µ
The problem under consideration has a very strong motivation and also is of independent
interest. In [2] it is an important ingredient of the proof of the following theorem. By Hs
we denote the s-dimensional Hausdorff measure.
Theorem 1.2 ([2]). Let d−1 < s < d, and let µ be a positive finite Borel measure such that
Hs(suppµ) < ∞. Then kRsµkL∞(md) = ∞ (equivalently, kRµsk = ∞).
If s is integer, the conclusion of Theorem 1.2 is incorrect. For 0 < s < 1 Theorem 1.2
was proved by Prat [10] using different approach. The obstacle for extension of this result to
all noninteger s between 1 and d−1 is the lack of the maximum principle. The same issue
concerns the quantitative version of Theorem 1.2 obtained by Jaye, Nazarov, and Volberg
[3].
The maximum principle is important for other problems on the connection between geo-
metric properties of a measure and boundedness of the operator Rs on L2(µ) – see for
µ
example [3], [5], [6], [7]. All these results are established for d−1 < s < d or s = d−1.
The problem of the lower estimate for kRsk in terms of the Wolff energy (a far going
µ
development of Theorem 1.2) which is considered in [3], [5], was known for 0 < s < 1. And
the results in [6], [7] are (d −1)-dimensional analogs of classical facts known for s = 1 (in
particular, [7] contains the proof of the analog of the famous Vitushkin conjecture in higher
dimensions). For 0 < s ≤ 1, the proofs essentially use the Melnikov curvature techniques
and do not require the maximum principle. But this tool is absent for s > 1.
At the same time the validity of the maximum principle itself remained open even for
0 < s < 1. It is especially interesting because the analog of (1.1) does not hold for each
component of Rsµ when 0 < s < d−1 unlike the case d−1 ≤ s < d – see Proposition 2.1
below.
Weprove Conjecture 1.1 for0 < s < 1in Section2 (Theorem 2.3). The proofiscompletely
different from the proof in the case d − 1 ≤ s < d. In Section 3 we prove Conjecture 1.1
in the special case of radial density of µ (that is when dµ = h(|x|)dm (x)), but for all
d
s ∈ (0,d). Section 4 contains an example showing that Conjecture 1.1 is incorrect for non-
positivemeasures, evenforradialmeasureswithC∞-density(notethatin[14,Conjecture7.3]
Conjecture 1.1 was formulated for all finite signed measures with compact support and C∞-
density).
2. The case 0 < s < 1
We start with a statement showing that the maximum principle fails for every component
of Rsµ if 0 < s < d−1.
ON THE MAXIMUM PRINCIPLE FOR THE RIESZ TRANSFORM 3
Proposition 2.1. For any d ≥ 2, 0 < s < d−1, and any M > 0, there is a positive measure
µ in Rd with C∞-density such that
sup |Rsµ(x)| > M sup |Rsµ(x)|, (2.1)
1 1
x∈Rd x∈suppµ
where Rsµ is the first component of Rsµ.
1
Proof. Let E = {(x ,...,x ) ∈ Rd : x = 0, x2 + ··· + x2 ≤ 1}, and let E , δ > 0, be
1 d 1 2 d δ
a δ-neighborhood of E in Rd. Let µ = µ be a positive measure supported on E with
δ δ
µ(E ) = 1 and with C∞-density ρ(x) such that ρ(x) < 2/vol(E ) ≤ C /δ. Then
δ δ d
|Rsµ(x′)| > A , where x′ = (1,0,...,0), 0 < δ < 1/2.
1 d
On the other hand, for x ∈ suppµ integration by parts yields
1 δ
|Rsµ(x)| < dµ(y)+ dµ(y)
1 Z |y −x|s Z |y −x|s+1
|y−x|<δ |y−x|≥δ
µ(B(x,δ)) δ µ(B(x,r)) ∞ µ(B(x,r))
= +s dr +δ(s+1) dr
δs Z rs+1 Z rs+2
0 δ
C δd Cs δ rd Cδ(s+1) 2 rd−1δ
d
< C + dr + dr +Cδ.
δ δs δ Z rs+1 δ Z rs+2
0 δ
Here by C we denote different constants depending only on d, and B(x,r) := {y ∈ Rd :
|y −x| < r}. We have
2
2 rd−1 δln , s = d−2,
δ dr =  δ
1
Zδ rs+2  (2d−s−2δ −δd−s−1), s 6= d−2.
d−s−2
Thus, all terms in the right-hand side of the estimate for |Rsµ(x)| tend to 0 as δ → 0, and
1
we may choose δ and a corresponding measure µ satisfying (2.1). (cid:3)
We need the following lemma. The notation A ≈ B means that cA < B < CB with
constants c,C which may depend only on d and s.
Lemma 2.2. Let µ be a non-negative measure in Rd with continuous density and compact
support. Let 0 < s < d−1. Then for every ball B = B(x ,r),
0
∞ dµ(B(x ,t))
(Rsµ·n)dσ ≈ rd−s−1µ(B)+rd 0 , (2.2)
(cid:12)Z (cid:12) Z ts+1
(cid:12) ∂B (cid:12) r
where n is the oute(cid:12)r normal vector(cid:12)to B and σ is the surface measure on ∂B.
(cid:12) (cid:12)
Proof. We will use the Ostrogradsky-Gauss Theorem and differentiation under the integral
sign. To justify these operations and make an integrand sufficiently smooth, we approximate
K(x) = x/|x|s+1 by the smooth kernel K in the following standard way. Let φ(t), t ≥ 0, be
ε
a C∞-function such that φ(t) = 0 as 0 ≤ t ≤ 1, φ(t) = 1 as t ≥ 2, and 0 ≤ φ′(t) ≤ 2, t > 0.
Let φ (t) := φ(t), K (x) := φ (|x|)K(x), and Rsµ := K ∗µ. We have
ε ε ε ε ε ε
(Rsµ·n)dσ = ∇·Rsµ(x)dm (x) = e ∇·φ (|y −x|) y −x dµ(y) dm (x).
Z∂B ε ZB ε d ZB(cid:20)ZRd ε |y−x|s+1 (cid:21) d
e e
The inner integral is equal to
y −x y −x
∇·φ (|y −x|) dµ(y)+ ∇· dµ(y) =: I (x)+I (x).
Z ε |y −x|s+1 Z |y −x|s+1 1 2
|y−x|≤2ε |y−x|>2ε
4 VLADIMIREIDERMANAND FEDOR NAZAROV
One can easily see that
∂ y −x 1 1 1 C
φ (|y −x|) < C + < , |y −x| ≤ 2ε.
ε
(cid:12)∂x (cid:20) |y −x|s+1(cid:21)(cid:12) (cid:20)ε|y −x|s |y −x|s+1(cid:21) |y−x|s+1
(cid:12) i (cid:12)
(cid:12) (cid:12)
Hence,
(cid:12) (cid:12)
1 2ε 1
|I (x)| < C dµ(y) < C dµ(B(x,t))
1 Z |y −x|s+1 Z ts+1
|y−x|≤2ε 0
µ(B(x,2ε)) 2ε µ(B(x,t))
≈ + dt.
(2ε)s+1 Z ts+2
0
Since µ has a continuous density with respect to m , we have µ(B(x,t)) < A td as t ≤ 2ε <
d µ,B
1, x ∈ B. Taking into account that s < d−1, we obtain the relation I (x)dm (x) → 0
B 1 d
as ε → 0. R
To estimate the integral of I (x) we use the equality ∇· x = d−s−1. Thus,
2 |x|s+1 |x|s+1
dµ(y)
I (x)dm (x) = C dm (x)
(cid:12)Z 2 d (cid:12) Z (cid:20)Z |y −x|s+1(cid:21) d
(cid:12) B (cid:12) B |y−x|>2ε
(cid:12) (cid:12)
(cid:12) (cid:12) dmd(x)
= C dµ(y)
(cid:18)Z (cid:20)Z |y−x|s+1(cid:21)
B(x0,r+2ε) B∩{|y−x|>2ε}
dm (x)
d
+ dµ(y) =: C(J +J ).
ZRd\B(x0,r+2ε)(cid:20)ZB |y−x|s+1(cid:21) (cid:19) 1 2
Obviously,
r td−1dt
≈ rd−s−1, |y−x | ≤ r,
dmd(x) Z ts+1 0
≈ 0
ZB |y −x|s+1  rd , |y−x | > r.
|y −x |s+1 0
0

In order to estimate J1 we note that for sufficiently small ε,
dm (x) dm (x)
d ≈ d ≈ rd−s−1, y ∈ B(x ,r +2ε).
Z |y −x|s+1 Z |y −x|s+1 0
B∩{|y−x|>2ε} B
Hence, J ≈ rd−s−1µ(B(x ,r+2ε)). Moreover,
1 0
rd ∞ dµ(B(x ,t))
J ≈ dµ(y) = rd 0 .
2 ZRd\B(x0,r+2ε) |y −x0|s+1 Zr+2ε ts+1
Passing to the limit as ε → 0, we get (2.2) (cid:3)
Now we are ready to prove our main result.
Theorem 2.3. Let µ be a non-negative measure in Rd with continuous density and compact
support. Let 0 < s < 1. Then (1.1) holds with a constant C depending only on d and s.
Proof. Let us sketch the idea of proof. Let a measure µ be such that µ(B(y,t)) ≤ Cts,
y ∈ Rd, t > 0. For Lipschitz continuous compactly supported functions ϕ, ψ, define the form
hRs(ψµ),ϕi by the equality
µ
1 y −x
hRs(ψµ),ϕi = (ψ(y)ϕ(x)−ψ(x)ϕ(y))dµ(y)dµ(x);
µ 2 ZZRd×Rd |y −x|s+1
the double integral exists since |ψ(y)ϕ(x)−ψ(x)ϕ(y)| ≤ C |x−y|. If we assume in addition
ψ,ϕ
that ψdµ = 0, we may define hRs(ψµ),ϕi for any (not necessarily compactly supported)
µ
R
ON THE MAXIMUM PRINCIPLE FOR THE RIESZ TRANSFORM 5
bounded Lipschitz continuous function ϕ on Rd; here we follow [4]. Let suppψ ∈ B(0,R).
For |x| > 2R we have
y −x y −x x
ψ(y)dµ(y) = + ψ(y)dµ(y)
(cid:12)ZRd |y −x|s+1 (cid:12) (cid:12)ZRd (cid:20)|y −x|s+1 |x|s+1(cid:21) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) C C (cid:12)
(cid:12) (cid:12)≤ (cid:12) |yψ(y)|dµ(y)= ψ . (cid:12)
|x|s+1 ZRd |x|s+1
Choose a Lipschitz continuous compactly supported function ξ which is identically 1 on
B(0,2R). Then we may define the form hRs(ψµ),ϕi as
µ
y −x
hRs(ψµ),ϕi = hRs(ψµ),ξϕi + ψ(y)dµ(y) (1−ξ(x))ϕ(x)dµ(x).
µ µ ZRd(cid:20)ZRd |y −x|s+1 (cid:21)
The repeated integral is well defined because
dµ(x) ∞ µ(B(0,t)) ∞ 1
≤ C dt ≤ C dt.
Z |x|s+1 Z ts+2 Z t2
|x|>2R 2R 2R
Assuming that Theorem 2.3 is incorrect and using the Cotlar inequality we establish the
existence ofapositive measureν such thatν hasnopointmasses, theoperatorRs isbounded
ν
on L2(ν), and hRs(ψν),1i = 0 for every Lipschitz continuous function ψ with ψdµ = 0.
ν
It means that ν is a reflectionless measure, that is a measure without point mRasses with
the following properties: Rs is bounded on L2(ν), and hRs(ψν),1i = 0 for every Lipschitz
ν ν
continuous compactly supported function ψ such that ψdµ = 0. But according to the
recent result by Prat and Tolsa [11] such measures do nRot exist for 0 < s < 1. We remark
that the proof of this result contains estimates of an analog of the Melnikov’s curvature of a
measure. This is the obstacle to extent the result to s ≥ 1. We now turn to the details.
Suppose that C satisfying (1.1) does not exists. Then for every n ≥ 1 there is a positive
measure µ such that
n
1
sup |Rsµ (x)| = 1, sup |Rsµ (x)| ≤ .
n n
n
x∈Rd x∈suppµn
Let
µ(B(x,r))
θ (x,r) := , θ := supθ (x,r).
µ rs µ µ
x,r
We prove that
0 < c < θ < C. (2.3)
µn
The estimate from above is a direct consequence of Lemma 2.2. Indeed, for any ball B(x,r)
(2.2) implies the estimate
c rd−1 ≥ (Rsµ ·n)dσ ≥ Crd−s−1µ (B),
d n n
(cid:12)Z (cid:12)
(cid:12) ∂B (cid:12)
(cid:12) (cid:12)
which implies the desired inequa(cid:12)lity. (cid:12)
The estimate from below follows immediately from a Cotlar-type inequality
sup |Rsµ (x)| ≤ C sup |Rsµ (x)|+θ
n n µn
x∈Rd (cid:2)x∈suppµn (cid:3)
(see [8, Theorem 7.1] for a more general result).
6 VLADIMIREIDERMANAND FEDOR NAZAROV
Let B(x ,r ) be a ball such that θ (x ,r ) > c = c(s,d), and let ν (·) = r−sµ (x +r ·).
n n µn n n n n n n n
Then
x−x
Rsµ (x) = Rsν n , θ (y,t) = θ (r y +x ,r t).
n n(cid:18) r (cid:19) νn µn n n n
n
In particular, ν (B(0,1)) = θ (x ,r ) > c. Choosing a weakly converging subsequence of
n µn n n
{ν }, we obtain a positive measure ν. If we prove that
n
(a) ν(B(y,t)) ≤ Cts,
(b) hRsν,ψi = 0 for every Lipschitz continuous compactly supported function ψ with
ν
ψdν = 0,
R (c) the operator Rs is bounded on L2(ν),
ν
then ν is reflectionless, and we come to contradiction with Theorem 1.1 in [11] mentioned
above. Thus, the proof would be completed.
The property (a) follows directly from (2.3). For weakly converging measures ν with
n
θ < C we may apply Lemma 8.4 in [4] which yields (b). To establish (c) we use the
µn
inequality
Rs,∗µ(x) := sup|Rµs,ε1(x)| ≤ kRsµkL∞(md) +C, x ∈ Rd, C = C(s),
ε>0
for any positive Borel measure µ such that µ(B(x,r)) ≤ rs, x ∈ Rd, r > 0, – see [12,
Lemma 2] or [13, p. 47], [1, Lemma 5.1] for a more general setting. Thus, Rsν (x) :=
ε n
Rs 1(x) ≤ C for every ε > 0. Hence, Rsν(x) ≤ C for ε > 0, x ∈ Rd, and the non-
νn,ε ε
homogeneous T1-theorem [9] implies the boundedness of Rs on L2(ν). (cid:3)
ν
3. The case of radial density
Lemma 2.2 allows us to prove the maximum principle for all s ∈ (0,d) in the special case
of radial density.
Proposition 3.1. Let dµ(x) = h(|x|)dm (x), where h(t) is a continuous function on [0,∞),
d
and let s ∈ (0,d−1). Then (1.1) holds with a constant C depending only on d and s.
Weremindthatfors ∈ [d−1,d)Conjecture1.1isprovedin[2]foranycompactlysupported
measure with C∞ density. Thus, for compactly supported radial measures with C∞ density
(1.1) holds for all s ∈ (0,d).
Proof. Because µ is radial, by (2.2) we have
c rd−1|Rsµ(x)| = (Rsµ·n)dσ
d
(cid:12)Z (cid:12)
(cid:12) ∂B(0,r) (cid:12)
(cid:12) (cid:12) ∞ dµ(B(0,t))
≈ (cid:12)rd−s−1µ(B(0,r))+r(cid:12)d , r = |x|.
Z ts+1
r
Thus,
µ(B(0,r)) ∞ dµ(B(0,t))
|Rsµ(x)| ≈ +r .
rs Z ts+1
r
Fix w 6∈ suppµ, and let r = |w|. If
µ(B(0,r)) ∞ dµ(B(0,t))
≥ r ,
rs Z ts+1
r
ON THE MAXIMUM PRINCIPLE FOR THE RIESZ TRANSFORM 7
then there is r ∈ (0,r) such that {y : |y| = r } ⊂ suppµ and µ(B(0,r)) = µ(B(0,r )).
1 1 1
Hence,
µ(B(0,r)) µ(B(0,r ))
|Rsµ(w)| ≈ < 1 ≤ C|Rsµ(x )|, |x | = r .
rs rs 1 1 1
1
If
µ(B(0,r)) ∞ dµ(B(0,t))
< r ,
rs Z ts+1
r
then there is r > r such that {y : |y| = r } ⊂ suppµ and µ(B(0,r)) = µ(B(0,r )). Hence,
2 2 2
µ(B(0,r )) µ(B(0,r)) ∞ dµ(B(0,t))
2
< < r ,
rs rs Z ts+1
2 r2
and we have
∞ dµ(B(0,t))
|Rsµ(w)| ≈ r ≤ C|Rsµ(x )|, |x | = r .
Z ts+1 2 2 2
r2
(cid:3)
4. Counterexample
Given ε > 0, we construct a signed measure ν = ν(ε) in R5 with the following properties:
(a) ν is a radial signed measure with C∞-density;
(b) suppν ∈ D := {1−ε ≤ |x| ≤ 1+ε};
ε
(c) |R2ν(x)| < ε for x ∈ suppν; |R2ν(x)| > a > 0 for |x| = 2, where a is an absolute
constant. Here R2ν means Rsν with s = 2.
Let ∆2 := ∆◦∆, and let
2/3, |x| ≤ 1,
u(x) =  1 1
− , |x| > 1.

|x| 3|x|3
Note that ∆( 1 ) = 0 and ∆2( 1 ) = 0in R5 \{0}. Hence, ∆2u(x) = 0,|x| =6 1. Moreover,
|x|3 |x|
∇u is continuous in R5 and ∇u(x) = 0, |x| = 1.
For δ ∈ (0,ε), let ϕ (x) be a C∞-function in R5 such that ϕ > 0, suppϕ = {x ∈
δ δ δ
R5 : |x| ≤ δ}, and ϕ (x)dm (x) = 1 (for example, a bell-like function on |x| ≤ δ}). Let
δ 5
U := u ∗ ϕ . ThenR∆2U (x) = 0 as x 6∈ D . Also, ∆U (x) → 0 as |x| → ∞. Hence, the
δ δ δ δ δ
function ∆U can be represented in the form ∆U = c( 1 ∗∆2U ) (here and in the sequel by
δ δ |x|3 δ
c we denote various absolute constants). Set dν = ∆2U dm . Then suppν ∈ D , and (b)
δ δ 5 δ δ
is satisfied. Since ∆( 1 ) = c , we have ∆U = c∆( 1 ∗∆2U ), that is U = c( 1 ∗∆2U )+h,
|x| |x|3 δ |x| δ δ |x| δ
where h is a harmonic function in R5. Since both U and 1 ∗∆2U tend to 0 as x → ∞, we
δ |x| δ
have
1
U = c ∗∆2U .
δ δ
(cid:18)|x| (cid:19)
Thus,
y −x
R2ν(x) = c dν (y) = c∇U (x).
δ δ
Z |y −x|3
Obviously, ∇U = ∇(u∗ϕ ) = (∇u)∗ϕ , and hence max |∇U (x)| → 0 as δ → 0. On the
δ δ δ x∈Dδ δ
other hand, for fixed x with |x| > 1 (say, |x| = 2) we have lim |∇u∗ϕ | = |∇u(x)| > 0.
δ→0 δ
Thus, (c) is satisfied if δ is chosen sufficiently small.
8 VLADIMIREIDERMANAND FEDOR NAZAROV
Remark. It is well-known that the maximum principle (with a constant C) holds for
potentials K(|x−y|)dµ(y) with non-negative kernels K(t) decreasing on (0,∞), and non-
negative finRite Borel measures µ. Our arguments show that for non-positive measures the
analog of (1.1) fails even for potentials with positive Riesz kernels. In fact we have proved
that for every ε > 0, there exists a signed measure η = η(ε) in R5 with C∞-density and such
that suppη ∈ D := {1−ε ≤ |x| ≤ 1+ε}, |u (x)| < ε for x ∈ suppη, but |u ((2,0,...,0))| >
ε η η
b > 0, where u (x) := dη(y), and b is an absolute constant.
η |y−x|
Indeed, for the firstRcomponent R2ν of R2ν we have
1
∂ 1 ∂ ∂
R2ν = c U = c ∗ (∆2U ) = cu , where dη = (∆2U )dm .
1 ∂x δ (cid:18)|x| ∂x δ (cid:19) η ∂x δ 5
1 1 1
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Vladimir Eiderman, Department of Mathematics, Indiana University, Bloomington, IN
E-mail address: [email protected]
Fedor Nazarov, Department of Mathematics, Kent State University, Kent, OH
E-mail address: [email protected]